Question 14 Marks
A cuboidal vessel is $10m$ long and $8m$ wide. How high must it be made to hold $380$ cubic meters of a liquid?
AnswerGiven that:
Length of the vessel $(I)=10 m$
Width of the Cuboidal vessel $=8 m$
Let ' $h$ ' be the height of the cuboidal vessel.
Volume of the vessel $=380 m^3$
Therefore, $l \times b \times h =380 m^3$
$\Rightarrow 10 \times 8 \times h=380$
$\Rightarrow h=\frac{380}{10 \times 8}$
$\Rightarrow h=4.75 m$
$\therefore$ Height of the vessel should be $4.75$ m .
View full question & answer→Question 24 Marks
A rectangular tank is $80\ m$ long and $25\ m$ broad. Water flows into it through a pipe whose cross-section is $25\ cm^2$, at the rate of $16\ km$ per hour. How much the level of the water rises in the tank in $45$ minutes?
AnswerConsider $'h'$ be the rise in water level.
Volume of water in rectangular tank $= 8000 \times 2500 \times h cm^2$
Cross-sectional area of the pipe $= 25\ cm^2$
Water coming out of the pipe forms a cuboid of base area $25\ cm^2$ and length equal to the distance travelled in $45$ minutes with the speed $16\ km/hour$ i.e.,
length = Length $=16000\times100\times\frac{45}{60}\text{cm}$
Therefore, The Volume of water coming out pipe in $45$ minutes $=25\times16000\times100\times\Big(\frac{45}{60}\Big)$
Now, volume of water in the tank = Volume of water coming out of the pipe in $45$ minutes $\Rightarrow8000\times2500\times\text{h}=16000\times100\times\frac{45}{60}\times25$
$\Rightarrow\text{h}=\frac{25\times16000\times100\times45}{60\times8000\times2500}=1.5\text{cm}$
View full question & answer→Question 34 Marks
Hameed has built a cubical water tank with lid for his house, with each other edge $1.5\ m$ long. He gets the outer surface of the tank excluding the base, covered with square tiles of side $25\ cm$ . Find how much he would spend for the tiles if the cost of tiles is $₹ 360$ per dozen.
AnswerGiven that: Hameed is getting $5$ outer faces of the tank covered with tiles,
he would need to know the surface area of the tank, to decide on the quantity of tiles required.
Edge of the cubical tank $(a) = 1.5m = 150cm$
So, surface area of the tank $= 5 \times 150 \times 150cm^2$
Area of each square tile $=\frac{\text{Surface Area of Tank}}{\text{Area of each Tile}}$
$=\frac{5\times150\times150}{25\times25}$
$= 180$
Cost of $1$ dozen tiles, i.e., cost of $12$ tiles $= ₹ 360$
$\therefore$ Cost of one tile $=\frac{₹ \ 360}{12}=₹\ 30$
So, the cost of $180 tiles = 180 \times 30 = ₹ 5400.$
View full question & answer→Question 44 Marks
The paint in a certain container is sufficient to paint on an area equal to $9.375\ m^2$, How many bricks of dimension $22.5\ cm \times 10\ cm$ $\times 7.5\ cm$ can be painted out of this container?
AnswerThe paint in the container can paint the area,
$A=9.375 m^2=93750 cm^2$ [Since $1 m=100\ cm$ ]
Dimensions of a single brick, Length $( l )=22.5\ cm$
Breadth $(b)=10\ cm$
Height $(h)=7.5\ cm$
We need to find the number of bricks that can be painted.
Surface area of $a^2$ brick $A^{\prime}=2( lb$ $+b h+h l)$.
$=2(22.5 \times 10+10 \times 7.5+7.5 \times 22.5)$
$=2(225+75+168.75)$
$=937.50 cm^2$
Number of bricks that can be painted
$=\frac{A}{A^{\prime}}$ $=\frac{93750}{937.5}=100$
Hence $100$ bricks can be painted out of the container.
View full question & answer→Question 54 Marks
A cuboidal water tank is $6m$ long, $5m $wide and $4.5m$ deep. How many liters of water can it hold?
AnswerGiven data:
Length $( l )=6 m$
Breadth $(b)$ $=5 m$
Height $( h )=4.5 m$
Volume of the tank $= I \times b \times h$
$= 6 \times 5 \times 4.5$
$= 135m^3$
It is given that,
It is given that,
$1 m^3=1000 \text { liters }$
Therefore, $135 m^3=(135 \times 1000)$ liters $=135000$ liters
The tank can hold $1,35,000$ liters of water.
View full question & answer→Question 64 Marks
A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in Figure. If the edge of each cube is $3\ cm$, find the volume of the structure built by the child.
AnswerVolume of each cube
$=$edge $\times$ edge $\times$ edge
$=3 \times 3 \times 3=27 cm^3$
Number of cubes in the structure $=15$
Therefore, volume of the structure $=27 \times 15=405\ cm^3$
View full question & answer→Question 74 Marks
Ravish wanted to make a temporary shelter for his car by making a box-like structure with the tarpaulin that covers all the four sides and the top of the car (with the front face of a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how many tarpaulins would be required to make the shelter of height $2.5\ m$ with base dimensions $4m \times 3m?$
AnswerGiven that: Shelter length $= 4m$
Breadth $= 3m$
Height $= 2.5m$
The tarpaulin will be required for top and four sides of the shelter.
The Area of tarpaulin required
$= 2h(l + b) + lb$
$\Rightarrow 2 \times 2.5 (4 + 3) + 4 \times 3$
$\Rightarrow 5(7) + 12$
$\Rightarrow 47m^2$
View full question & answer→Question 84 Marks
The length of a hall is $18m$ and the width $12m$. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.
AnswerLength of the hall $= 18\ m$
Width of the hall $= 12\ m$
Now given, Area of the floor and the flat roof = sum of the areas of four walls
$\Rightarrow 2 \times lb = 2 \times lh + 2 \times bh $
$\Rightarrow lb = lh + bh$
$\Rightarrow\text{h}=\frac{\text{lb}}{\text{l}+\text{b}}$
$=\frac{18\times12}{18\times12}$
$=\frac{216}{30}=7.2\text{m}$
View full question & answer→Question 94 Marks
Water in a canal $30\ dm$ wide and $12\ dm$ deep, is flowing with a velocity of $100\ km$ every hour. What much area will it irrigate in $30$ $k$ minutes if $8\ cm$ of standing water is desired?
AnswerGiven that: Water in the canal forms a cuboid of
Width $(b) = 30 dm = 3m$
Height $(h) = 12dm = 1.2m$
Cuboid length is equal to the distance traveled in $30$ min with the speed of $100\ km$ per hour.
Therefore, Length of the cuboid
$=100\times\frac{30}{60}=60\text{km}=50000 \text{meters}$
So, volume of water to be used for irrigation $= 5000 \times 3 \times 1.2m^3$
Water accumulated in the field forms a cuboid of base area equal to the area of the field and height equal to $\frac{8}{100}\text{metres}$
Therefore, Area of field$\times\frac{8}{100}=50000\times3\times1.2$
$\Rightarrow\text{Area of field}=\frac{50000\times3\times1.2\times100}{8}$
$\Rightarrow 22,50,000$ metres.
View full question & answer→Question 104 Marks
A godown measures $40m \times 25m \times 10m$. Find the maximum number of wooden crates each measuring $1.5m \times 1.25m \times 0.5m$ that can be stored in the godown.
AnswerGiven: Godown length $\left(l_1\right)=40 m$
Godown breadth $\left(b_1\right)=25 m$
Godown height $\left(h_1\right)=10 m$
Volume of the godown $=$ $l _1 \times b _1 \times h _1=40 \times 25 \times 10=10000 m^3$
Wooden crate length $\left( l _2\right)=1.5 m$
Wooden crate breadth $\left( b _2\right)=1.25 m$
Wooden crate height $\left( h _2\right)=0.5 m$
Volume of the wooden crate $= I _2 \times b _2 \times h _2=1.5 \times 1.25 \times 0.5=0.9375 m^3$
The number of wooden crates stored in the godown is taken as ' $n$ ' Volume of ' $n$ ' wooden crates
$=$ Volume of godown $=$ $0.9375 n =10000= n =\frac{10000}{0.9375}=10666.66$
$\therefore$ The number of wooden crates that can be stored in the godown is $10666.66.$
View full question & answer→Question 114 Marks
The length, breadth, and height of a room are $5 m, 4 m$ and $3 m$ respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of $₹ 7.50 m^2$.
AnswerTotal Area to be washed $= lb +2( l + b ) h$
Where, length $( I )=5 m$
breadth (b) $=4 m$
height $(h)=3 m$
Therefore, the total area to be white washed is $=(5 \times 4)+2 \times(5+4) \times 3$ $=74 m^2$
Now, The cost of white washing $1 m^2$ is $₹ 7.50$
Therefore, the cost of white washing $74 m^2=(74 \times 7.50)$
$=₹ 555 /-$
View full question & answer→Question 124 Marks
The dimentional of rectangular box are inthe ratio of $2: 3: 4$ and the difference between the cost of covering it with sheet of paper at the rates of $₹ 8$ and $₹ 9.50$ per $m ^2$ is ₹ $1248 $. Find the dimensions of the box.
AnswerThe dimensions of the rectangular box are in the ratio $2: 3: 4$.
So, let the dimensions be, Lenght $(I)=(4 x) m$
Breadth $(b) =(3 x ) m$
Height $( h )=(2 x ) m$
We are asked to find the dimensions of the box The total surface area of the box, $A =$
$2( lb + bh + hl )=2[(4 x )(3 x )+(3 x )(2 x )+(2 x )(4 x )]=\left(52 x ^2\right) m ^2$
The cost of covering it at the rate of $₹ 8$ per $m ^2=₹(8 \times$ A)
The cost of covering it at the rate of $₹ 9.50$ per $m ^2=₹(9.50 \times A )$
We know that, The difference between above two costs is $₹ 1248$ .
So, $1248=(9.50) A -(8) A =1.50 A A =832 m^2 52 x ^2=832\left\{\right.$ Since $\left.A =52 x ^2\right\} x ^2=\frac{832}{52}=\frac{64}{4}=16 x$
$=4$
So, the dimensions of the box are; $2 x =2 \times 4=8 m 3 x =3 \times 4=12 m 4 x =4 \times 4=16 m$
Hence, The dimensions of the box are $8\ m, 12\ m$ and $16\ m$ .
View full question & answer→Question 134 Marks
The dimensions of a room are $12.5 m$ by $9 m$ by $7 m$ . There are $2$ doors and $4$ windows in the room; each door measures $2.5 m$ by $1.2 m$ and each window $1.5 m$ by $1 m $. Find the cost of painting the walls at $₹ 3.50$ per square meter.
AnswerGiven Length of the room $=12.5 m$
Breadth of the room $=9 m$
Height of the room $=7 m$
$\therefore$ Total surface area of the four walls $=2(1+b) \times h$
$=2(12.5+9) \times 7$
$=301 m^2$
Area of $2$ doors $=2(2.5 \times 1.2)$
$=6 m^2$
Area of $4$ windows $=4(1.5 \times 1)$
$=6 m^2$
Area to be painted on $4$ walls $=301-(6+6)$
= $301-12$
$=289 m^2$
Therefore, Cost of painting $=289 \times 3.50$
$= ₹ 1011.5$
View full question & answer→Question 144 Marks
A closed iron tank $12\ m$ long, 9m wide and $4 \ m$ deep is to be made. Determine the cost of iron sheet used at the rate of ₹ $5$ per meter sheet, a sheet being $2\ m$ wide.
AnswerGiven that: Length $(I)=12 m$
Breadth $(b)=9 m$
Height $(h)=4 m$
Total surface area of the tank $=2[l b+b h+h l]=2[12 \times 9+9 \times 4+12 \times 4]=2[108+36+48]=384 m^2$
The lenght of the Iron sheet $=\frac{\text { Area of the Iron Sheet }}{\text { Width of the Iron Sheet }}=\frac{384}{2}=192 m$.
Cost of the Iron Sheet $=$ Length of the Iron Sheet $\times$ Cost rate $=192 \times 5=₹ 960$
View full question & answer→Question 154 Marks
An open box is made of wood $3\ cm$ thick. Its external length, breadth and height are $1.48\ m$, $1.16\ m$ and $8.3\ dm$. Find the cost of painting the inner surface of ₹ $50$ per sq. metre.
AnswerGiven that:
Outer DimensionsLength $=148\ cm$
Breadth $=116\ cm$
Height $=83\ cm$
Inner Dimensions Length $=148-(2 \times 3)=142 cm$
Breadth $=116-(2 \times 3)=110\ cm$
Height $=83-3=80$
Surface Area of the Inner region $=2 h( I + b )+ lb =2 \times 80(142+110)+142 \times 110=2 \times 80 \times 252+142 \times 110=$
$55940\ cm^2=5.2904\ m^2$
Hence, the cost of Painting the Surface Area of the Inner region $=5.2904 \times 50=₹ 279.70$
View full question & answer→Question 164 Marks
A wooden bookshelf has external dimensions as follows: Height $= 110\ cm$, Depth $= 25\ cm,$ Breadth $= 85\ cm$ (See figure). The thickness of the plank is $5\ cm$ everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is $20$ paise per $cm^2$. Find the total expenses required for polishing and painting the surface of the bookshelf.
AnswerExternal length of book shelf $=85 cm$
Breadth $=25 cm$
Height $=110 cm$
External surface area of shelf while leaving front face of shelf
$lh +2( lb + bh )[85 \times 110+2(85 \times 25+25 \times 110)]$ $19100 cm^2$
Area of Front face $=[85 \times 110-75 \times 100+2(75 \times 5)] cm ^2=1850+750 cm^2=2600 cm^2$
Area to be polished $=19100+2600 cm^2=21700 cm^2$
Cost of polishing $1 cm^2$ area $=₹ 0.20=₹ 4340$
Now, Length $(I)$, breadth $(b)$, height $(h)$ of each row of book shelf is $75\ cm, 20\ cm$ and $30\ cm$ $=\left(\frac{110-20}{3}\right)$ respectively.
Area to be painted in $1 row =2(1+ h ) b + lh [2(75+30) \times 20+75 \times 30)] cm ^2(4200+2250) cm ^2 6450 cm^2$
Area to be painted in $3$ rows $=3 \times 6450=₹ 1935\ cm^2$
View full question & answer→Question 174 Marks
A village having a population of $4000$ requires $150$ liters of water per head per day. It has a tank measuring $20\ m × 15\ m × 6\ m.$ For how many days will the water of this tank last?
AnswerGiven that: Length of the cuboidal tank $(l) = 20m$
Breadth of the cuboidal tank $(b) = 15m$
Height of the cuboidal tank $(h) = 6m$
Capacity of the tank $= l \times b \times h = 20 \times 15 \times 6 = 1800m^3 = 1800000$
litres Water consumed by the people of village in one day $= 4000 \times 150$ litres $= 600000$ litres
Let water of this tank last for $'n'$ days
Therefore, water consumed by all people of village in $n$ days = Capacity of the tank
$= n \times 600000 = 1800000$
$=\text{n}=\frac{1800000}{600000}=3$
Thus, the water will last for $3$ days in the tank.
View full question & answer→Question 184 Marks
Two cubes, each of volume $512\ cm^3$ are joined end to end. Find the surface area of the resulting cuboid.
AnswerGiven that: Volume of the cube $=512\ cm^3$
$\Rightarrow side^3= 512$
$\Rightarrow side^3 = 8^3$
$\Rightarrow side = 8\ cm$
Dimensions of the new cuboid formed Length $( l )=8+8=16 cm$, Breadth $(b) =8 cm$,
Height $( h )=8\ cm$
Surface area $=2(l b+b h+h l)=2(16 \times 8+8 \times 8+16 \times 8)=640 cm^2$
$\therefore$ Surface area is $640\ cm^2$.
View full question & answer→Question 194 Marks
The length and breadth of a hall are in the ratio $4 : 3$ and its height is $5.5$ meters. The cost of decorating its walls (including doors and windows) at ₹ $6.60$ per square meter is ₹ $5082$. Find the length and breadth of the room.
AnswerLet the length be $4$ a and breadth be $3$ a Height $=5.5 m$ [Given]
As mentioned in the question, the cost of decorating 4 walls at the rate of $Rs.\ 6.60$ per $m ^2$ is $₹ 5082$.
Area of four walls $\times$ rate $=$ Total cost of Painting $2(1+b) \times h \times 6.6$
$=50822(4 a+3 a) \times 5.5 \times 6.6$
$=5082$
$7 a=\frac{5082}{2 \times 5.5 \times 6.6} 7 a=70 a=\frac{70}{7} a=10$
Length $=4 a=4 \times 10=40\ m$ Breadth $=3 a =3 \times 10=30\ m$.
View full question & answer→Question 204 Marks
A box with lid is made of $2\ cm$ thick wood. Its external length, breadth and height are $25\ cm$, $18\ cm$ and $15\ cm$ respectively. How much cubic cm of a fluid can be placed in it? Also, find the volume of the wood used in it.
AnswerGiven: The external dimensions of cuboid are as follows:
Length $( l )=25 cm$
Breadth $( b )=18 cm$
Height $( h )=15 cm$
External volume of the case with cover (cuboid)
$= I \times b \times h\ cm ^3=25 \times 18 \times 15\ cm^3=6750 cm^3$
Now, the internal dimensions of the cuboid is as follows
Length $( l )=25-(2 \times 2)=21 cm$
Breadth $(b)=18-(2 \times 2)=14\ cm$
Height $(h)=15-(2 \times 2)=11\ cm$
Now, Internal volume of the case with cover (cuboid) $=1 \times b \times h\ cm ^3=21 \times 14 \times 11 cm^3=3234 cm^3$
Therefore, Volume of the fluid that can be placed $=3234\ cm^3$
Now, volume of the wood utilized $=$ External volume - Internal volume $=3516 cm^3$
View full question & answer→Question 214 Marks
The cost of preparing the walls of a room $12m$ long at the rate of $₹ 1.35$ per square meter is $₹ 340.20$ and the cost of matting the floor at $85$ paise per square meter is $₹ 91.80$. Find the height of the room.
AnswerGiven that:
Length of the room $= 12m$
Let the height of the room be $'h'$
Area of $4$ walls $= 2(l + b) \times h$
According to the question
$2(l + b) \times h \times 1.35 = 340.20$
$2(12 + b0 \times h \times 1.35) = 340.20$
$(12+\text{b})\times\text{h}=\frac{170.10}{1.35}=126\dots(1)$
Also Area of the Floor $= l \times b$
Therefore,$l \times b \times 0.85 = 91.80$
$\Rightarrow 12 \times b \times 0.85 = 91.80$
$\Rightarrow b = 9m ...(2)$
Substituting $b = 9m$ in equation $(1)$
$(12 + 9) \times h = 126; h = 6m.$
View full question & answer→Question 224 Marks
Half cubic meter of gold-sheet is extended by hammering so as to cover an area of $1$ hectare. Find the thickness of the gold-sheet.
AnswerGiven that: Volume of gold-sheet $= 0.5m^3$
Area of the gold-sheet $= 1 hectare = 1 \times 10000 = 10000 m^2$
Therefore, Thickness of gold sheet $=\frac{\text{Volume of solid}}{\text{Area of gold sheet}}$
$\Rightarrow\frac{0.5\text{m}^3}{1\text{ Hectare}}$
$\Rightarrow\frac{0.5\text{m}^3}{10000\text{m}^2}$
$\Rightarrow\frac{100\text{m}}{20000}$
$\therefore$ Thickness of silver sheet $=\frac{1}{200}\text{cm}$
View full question & answer→Question 234 Marks
Each edge of a cube is increased by $50\%$. Find the percentage increase in the surface area of the cube.
AnswerLet $'a'$ be the edge of the cube Therefore the surface area of the cube
$= 6a^2 I.e., S_1 = 6a^2$
We get a new edge after increasing the edge by $50 \%$ The new edge
$=\text{a}+\frac{50}{100}\times\text{a}$
$=\frac32\times\text{a}$
Considering the new edge, the new surface area is
$=6\times\Big(\frac32\text{a}\Big)^2$ I.e., $\text{S}_2=6\times\frac94\text{a}^2$
$\text{S}_2=\frac{27}{2}\text{a}^2$
$\therefore$ Increase in the Surface Area
$=\frac{27}{2}\text{a}^2-6\text{a}^2$
$=\frac{15}{2}\text{a}^2$
So, increase in the surface area
$=\frac{\frac{15}{2}\text{a}^2}{6\text{a}^2}\times100$
$=\frac{15}{12}\times100$
$=125\%$
View full question & answer→Question 244 Marks
The breadth of a room is twice its height, one half of its length and the volume of the room is $512cu$. dm. Find its dimensions.
AnswerConsider $I , b$ and h are the length, breadth and height of the room.
So, $b =2 h$ and $b =\left(\frac{1}{2}\right) l$
$\Rightarrow \frac{1}{2}=2 h$
$\Rightarrow l=4 h$
$\Rightarrow l=4 h, b=2 h$
Now, Volume $=512\ dm ^3$
$\Rightarrow 4 h \times 2 h \times h=512$
$\Rightarrow h^3=64$
$\Rightarrow h=4$
So, Length of the room $( l )=4 h=4 \times 4=16\ dm$
Breadth of the room $(b)=2 h=2 \times 4=8\ dm$ And Height of the room $( h )=4\ dm$.
View full question & answer→Question 254 Marks
Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as $80\ cm$, $40\ cm$ and $20\ cm$ respectively. How many square sheets of paper of side $40\ cm$ would she require?
AnswerGiven that: Mary wants to paste a paper on the outer surface of the wooden block.
The quantity of the paper required would be equal to the surface area of the box which is of the shape of a cuboid.
The dimensions of the wooden block are:
Length $( l )=80\ cm$
Breadth (b) $=40\ cm$
Height $( h )=20\ cm$
Surface Area of the wooden box $=2[ lb + bh + hl ]=2[(80 \times 40)+(40 \times 20)+(20 \times 80)]=2[5600]=11200 cm^2$
The Area of each sheet of the paper $=40 \times 40 cm^2=1600 cm^2$
$\therefore$ The number of sheets required $=\frac{\text { Surface area of the box }}{\text { Area of one sheet of paper }}=\frac{11200}{1600}=7$
So, she would require $7$ sheets.
View full question & answer→Question 264 Marks
If $V$ is the volume of a cuboid of dimensions $a, b, c$ and $S$ is its surface area, then prove that:
$\frac{1}{\text{V}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)$
AnswerGiven Data:
Length of the cube $(l) = a$
Breadth of the cube $(b) = b$
Height of the cube $(h) = c$
Volume of the cube $(V) = l \times b \times h = a \times b \times c = abc$
Surface area of the cube $(S) = 2 (lb + bh + hl) = 2(ab + bc + ca)$
Now, $\frac{\text{ab}+\text{bc}+\text{ca}}{\text{abc}}\frac{2}{2(\text{ab}+\text{bc}+\text{ca})}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)$
$\therefore\ \frac{1}{\text{abc}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)$
$\therefore\ \frac{1}{\text{V}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)$ Hence Proved.
View full question & answer→Question 274 Marks
A river $3m$ deep and $40\ m$ wide is flowing at the rate of $2\ km$ per hour. How much water will fall into the sea in a minute?
AnswerRadius of the water flow $= 2\ km$ per hour
$=\Big(\frac{2000}{60}\Big)\text{m/min}$
$=\Big(\frac{100}{3}\Big)\text{m/min}$
Depth of the river $(h) = 3\ m$
Width of the river $(b) = 40\ m$
Volume of the water flowing in $1$ min
$=\frac{100}{3}\times40\times3=4000^3$
Thus, $1$ minute $4000m^3 = 4000000$ litres of water will fall in the sea.
View full question & answer→Question 284 Marks
A wall of length $10\ m$ was to be built across an open ground. The height of the wall is $4\ m$ and thickness of the wall is $24\ cm$. If this wall is to be built up with bricks whose dimensions are $24cm \times 12cm \times 8cm$, how many bricks would be required?
AnswerGiven that: The wall with all its bricks makes up space occupies by it,
we need to find the volume of the wall, which is nothing but cuboid.
Here, length $= 10\ m = 1000\ cm$
Thickness $= 24\ cm$
Height $= 4\ m = 400\ cm$
Therefore, volume of the wall $= l \times b \times h = 1000 \times 24 \times 400cm^3$
Now, each brick is a cuboid with length $= 24\ cm$ Breadth $= 12\ cm$ Height $= 8\ cm$
So, volume of each brick $= l \times b \times h = 24 \times 12 \times 8 = 2304cm^3$
The number of bricks required is given by, $=\frac{\text{Volume of the wall}}{\text{Volume of each brick}}$
$=\frac{1000\times24\times400}{2304}=4166.6\text{ bricks}$
So, the wall requires $4167$ bricks.
View full question & answer→